Question

A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen. 1 litre of the hydrocarbon weighs 1.26 g at NTP. Determine the Molecular Formula of the Hydrocarbon.

Answer 1

Carbon M=m/mm m=2.40g/12=0.2mole

Hydrogen =0.5/1=0.5

Then our ratio should be 2:5

Our emperical element are C2H5

Answer 2

Answer:

The formula of hydrocarbon is

$C _{2} H _{4}$

Explanation:

Given that,

The percentage of carbon=85.7%

The percentage of Hydrogen=14.3%

The mole ratio of Carbon and hydrogen is as follows

$= \frac{85.7}{12} :14.3 = 7.14:14.3 \\ ≈ 1:2$

Now,the empirical formula of Hydrocarbon is written as

$C _{n}H _{2n}$

The formula weight of hydrocarbon is

$12 \times 1 + 2 \times 1 = 14$

Given that,1 litre of Hydrocarbon weighs 1.26g.Therefore,molecular weight is weight of 22.4litres of hydrocarbon i.e,

$= 22.4 \times 1.26 = 28.2$

Now we use the relation

$n = \frac{molecular \: wt}{formula \: wt}$

Therefore,

$n = \frac{28.2}{14} ≈2$

Hence the formula of Hydrocarbon becomes

$C _{2} H _{4}$

#SPJ3